Theo đề ra, ta có: \(x\inℤ\Leftrightarrow2x\inℤ\)
Ta có: \(2x+\frac{8}{5}-\frac{x}{5}=2x+\frac{\left(8-x\right)}{5}\)
Để \(L\inℤ\Leftrightarrow\frac{8-x}{5}\inℤ\)
\(\Leftrightarrow\left(8-x\right)⋮5\)
\(\Leftrightarrow\left(8-x\right)\in B\left(5\right)=\left\{x;\left|x=5g\right|g\inℤ\right\}\)
\(\Leftrightarrow\left(8-x\right)=5g\)
\(\Leftrightarrow x=8-5g\left(g\inℤ\right)\)